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Forum: Linear Algebra   Topic: Eigenvalues of Symmetric Matrices

 Post subject: Re: Eigenvalues of Symmetric Matrices Posted: Mon Dec 03, 2018 10:12 pm

Replies: 1
Views: 165

 This follows from the following characterisation of the eigenvalues of symmetric matrices. $\lambda_k(A) = \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^TAx}{\|x\|_2}$ Here the minimum is taken over all subspaces $U$ of $\mathbb{R}^n$ of dimension $n-k+1$. (I chose to write $\lambda_K(A)$ r...
 Posted: Tue Sep 26, 2017 10:09 am

Replies: 1
Views: 1349

 It remained unanswered for quite some time. Since I have been asked for it, here is a solution: The answer is affirmative. Given a sequence $a = (a_1,a_2,a_3,\ldots)$, where $a_i \in \{1,2\}$ for each $i$, we construct the subset $S_a$ of $(0,1)$ as follows: Every element of $S_a$ is of the form $... Forum: Analysis Topic: An interesting limit  Post subject: Re: An interesting limit Posted: Thu Jul 14, 2016 10:39 am Replies: 2 Views: 710  We have also seen this here : The basic steps are as follows: We consider independent Poisson distributions $X_1,X_2,\ldots$ with parameter 1. We know that $Y_n = X_1 + \cdots + X_n$ is Poisson with parameter $n$. From the central limit theorem, $(Y_n-n)/\sqrt{n}$ converges in distribution t... Forum: Real Analysis Topic: Inflection Point of a function  Post subject: Re: Inflection Point of a function Posted: Thu Jul 14, 2016 7:48 am Replies: 2 Views: 494  I presume you can show that $f'(1) = 1$ showing that $x=1$ is a stationary point. Now observe that for $0 < x < 1$ we have $x^x < x^1 = x$. So $x^{x^x} < x^x$, i.e. $f(x) < 0 = f(1).$ Also, for $x > 1$ we have $x^x > x$ giving $f(x) > f(1)$. So $x=1$ is a point of inflection. Forum: Real Analysis Topic: Infinite Series  Post subject: Re: Infinite Series Posted: Wed Jul 13, 2016 9:06 am Replies: 2 Views: 487  To use Stirling's approximation observe that \[ \prod_{r=1}^n \left(1 + \frac{r}{n}\right)^{1/n} = \left( \frac{(n+1)(n+2)\cdots (n+n)}{n^n}\right)^{1/n} = \left(\frac{(2n)!}{n^n(n!)}\right)^{1/n}.$ By Stirling's approximation we have $\frac{(2n)!}{n^n(n!)} \sim \frac{\sqrt{4\pi n}(2n/e)^{2n}}{n^... Forum: Calculus Topic: floor function integral  Post subject: Re: floor function integral Posted: Wed Jul 13, 2016 7:47 am Replies: 5 Views: 754  Indeed as Apostolos says, the integral diverges. What jacks's proof shows is that for every $0 < \varepsilon < \pi/2$ we have that $\int_{\varepsilon}^{\pi-\varepsilon} \lfloor \cot{x} \rfloor \; dx = -\pi + 2\varepsilon.$ However this does not mean that the value of the integral converges to$−... Forum: Real Analysis Topic: Positive function  Post subject: Re: Positive function Posted: Tue Jul 12, 2016 8:09 am Replies: 1 Views: 378  If this is a Riemann integral, we use the fact that any Riemann integrable function has a point of continuity. (In fact the set of point of discontinuity has measure zero - this is Lebesgue's criterion.) So there is an interval \(I \subseteq [0,x]$ and a positive real number $a$ such that $f(t) ... Forum: Complex Analysis Topic: Analytic Function  Post subject: Re: Analytic Function Posted: Tue Jul 12, 2016 7:36 am Replies: 1 Views: 556  No it is not. For example \(z$ is analytic but $\overline{z}$ is not. (As can be checked e.g. by the Cauchy-Riemann equations.)By the way, "analytic" and "locally represented by a convergent power series" mean the same thing. Forum: Real Analysis Topic: constant function  Post subject: Re: constant function Posted: Tue Jul 12, 2016 7:26 am Replies: 3 Views: 629  Hi Apostole,It is a nice proof! I have not seen it anywhere. Forum: Real Analysis Topic: constant function  Post subject: Re: constant function Posted: Tue Jul 12, 2016 7:25 am Replies: 3 Views: 629  Given two points $x,y \in \mathcal{D}$ we define \[\displaystyle{D(x,y) = \frac{f(x)-f(y)}{x-y}}.$ Note that for $x < y < z$ we have $D(x,y) = \alpha D(x,z) + (1-\alpha)D(z,y)$ where $\alpha = \alpha(x,y,z) = \displaystyle{\frac{z-x}{x-y} \in (0,1)}.$ It follows that either \(D(x,z) \leqsl...
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